In the previous project, a constant 12V DC power circuit with current limitation of 1A was designed. In this project, a symmetrical dual power supply with constant voltage outputs will be designed. A symmetrical dual power supply can provide two symmetrical voltages at the output with opposite polarity with respect to a common ground reference. Every electronic circuit needs a proper power supply at the input for its optimum functioning. The power supply of any device or circuit should be chosen as per its power requirements. In this project, a regulated power supply circuit which can output constant voltages of 9V and -9V with 1A maximum current is designed.
The power circuit designed in this project uses 7809 and 7909 voltage regulator IC and uses conventional steps of power circuit design like stepping down AC voltage, converting AC voltage to DC voltage and Smoothing DC voltage for getting direct input from the AC mains.
Components Required –
Fig. 1: List of Components required for Constant +/-9V DC Symmetrical Power Supply
Block Diagram –
Fig. 2: Block Diagram of Constant +/-9V DC Symmetrical Power Supply
Circuit Connections –
The circuit is built in stages with every stage aimed for a specific purpose. For stepping down the 230 V AC, a 12V – 0 -12V transformer is taken. The terminals of the secondary coil of the transformer are connected with a full-bridge rectifier and a wire is drawn from the center tape of the transformer to serve as common ground. The full bridge rectifier is built by connecting four 1N4007 diodes to each other designated as D1, D2, D3 and D4 in the schematics. The cathode of D1 and anode of D2 is connected to one of the secondary coil and cathodes of D3 and anode of D4 is connected to the other terminal of the coil. The cathodes of D2 and D4 are connected from which one terminal is taken out from output of rectifier and anodes of D1 and D3 are connected from which other terminal is taken out from output from full-wave rectifier.
A fuse of 1A is connected in series to the output of full-wave rectifier for protection from the AC supplies. The capacitors of 470 uF (shown as C1 and C2 in schematics) are connected between the output terminals of a full-wave rectifier for smoothing purpose. For voltage regulation LM-7809 and 7909 ICs are connected in parallel to the smoothing capacitors. The output is drawn from the voltage output terminals of the regulator ICs. The capacitors of 220 uF (shown as C3 and C4 in schematics) are connected to the output terminals of the power circuit to compensate transient currents.
How the circuit works –
The power circuit operates in stages with each stage serving a specific purpose. The circuit operates in the following stages –
1. AC to AC Conversion
2. AC to DC Conversion – Full Wave Rectification
4. Voltage Regulation
5. Compensating Transient Currents
6. Short Circuit Protection
AC to AC conversion
The voltage of Main Supplies (Electricity fed by the intermediate transformer after stepping down line voltage from generating station) is approximately 220-230V AC which further needs to be stepped down to 9V level. To reduce the 220V AC to 9V AC, a step-down transformer with center taping is used. The use of center tap transformer allows utilizing both positive and negative polarities of the voltage at the input. The circuit takes some drop in the output voltage due to resistive loss. Therefore a transformer of high voltage rating greater than the required 9 V needs to be taken. The transformer should provide 1A current at the output. The most suitable step-down transformer that meets the mentioned voltage and current requirements is 12V-0-12V/2A. This transformer step downs the main line voltage to +/-12V AC, as shown in the below image.
Fig. 3: Circuit Diagram of 12-0-12V Transformer
AC to DC conversion – Full Wave Rectification
The stepped down AC voltage needs to be converted to DC voltage through rectification. The rectification is the process of converting AC voltage to DC voltage. There are two ways to convert an AC signal to the DC one. One is half wave rectification and another is full wave rectification. In this circuit, a full-wave bridge rectifier is used for converting the 24V AC to 24V DC. The full wave rectification is more efficient than half wave rectification since it provides complete use of both the negative and positive sides of AC signal. In full wave bridge rectifier configuration, four diodes are connected in such a way that current flows through them in only one direction resulting in a DC signal at the output. During full wave rectification, at a time two diodes become forward biased and another two diodes get reverse biased.
Fig. 4: Circuit Diagram od Full Wave Rectifier
During the positive half cycle of the supply, diodes D2 and D3 conduct in series, while diodes D1 and D4 are reverse biased and the current, flows through the output terminal passing through D2, output terminal and the D3. During the negative half cycle of the supply, diodes D1 and D4 conduct in series, but diodes D3 and D2 are reverse biased and the current flows through D1, output terminal and the D4. The direction of current both ways through the output terminal in both conditions remain the same.
Fig. 5: Circuit Diagram showing positive cycle of Full Wave Rectifier
Fig. 6: Circuit Diagram showing negative cycle of Full Wave Rectifier
The 1N4007 diodes are chosen to build the full wave rectifier because they have the maximum (average) forward current rating of 1A and in reverse biased condition, they can sustain peak inverse voltage up to 1000V. That is why 1N4007 diodes are used in this project for full wave rectification.
Smoothing is the process of filtering the DC signal by using a capacitor. The output of the full-wave rectifier is not a steady DC voltage. The output of the rectifier has double the frequency of main supplies but it contains ripples. Therefore, it needs to be smoothed by connecting a capacitor in parallel to the output of full wave rectifier. The capacitor charges and discharges during a cycle giving a steady DC voltage as the output. So, 470 uF capacitors (shown as C1 and C2 in schematics) of high value are connected to the output of rectifier circuit. As the DC which is to be rectified by the rectifier circuit has many AC spikes and unwanted ripples, so to reduce these spikes capacitor is used. This capacitor acts as a filtering capacitor which bypasses all the AC through it to ground. At the output, the mean DC voltage left is smoother and ripple free.
Fig. 7: Circuit Diagram of Smoothing Capacitor
For providing a regulated +/-9V at the output, an LM-7809 and 7909 ICs are used. These ICs are capable of providing current up to 1A. The 7809 IC is a positive voltage regulator which gives stable +9V at the output with a positive input supply of 12V. For getting a negative voltage at the output a negative voltage regulator 7909 is used. It provides a -9V at the output on an input of -12V. The 7809 IC provides an output voltage in the range of 8.6V to 9.4V with the input voltage range of 11.5V to 24V while 7909 IC provides an output voltage in the range of -8.6V to -9.4V with the input voltage range of -11.5V to -23V. The common ground is provided by the center tap terminal of the transformer. Both of the regulator ICs are capable of load regulation themselves. They provide regulated and stabilized the voltage at the output irrespective of the fluctuation in the input voltage and load current.
The LM7809 and 7909 ICs have the following internally tolerable power dissipation –
Pout = (Maximum operating temperature of IC)/ (Thermal Resistance, Junction−to−Air + Thermal Resistance, Junction−to−case)
Pout = (125) / (65+5) (values as per the datasheet)
Pout = 1.78W
Therefore, both voltage regulator ICs can internally sustain up to 1.78W power dissipation. Above 1.78W, the ICs will not tolerate the amount of heat generated and will start burning. This can cause a serious fire hazard also. So heat sink is needed to dissipate the excessive heat from the ICs.
Compensating Transient Currents
At the output terminals of the power circuit, 220 uF capacitor ( shown as C3 and C4 in schematics) are connected in parallel. These capacitors help in fast response to load transients. Whenever the output load current changes then there is an initial shortage of current, which can be fulfilled by this output capacitor.
The output current variation can be calculated by
Output current ,Iout = C (dV/dt) where
dV = Maximum allowable voltage deviation
dt = Transient response time
Considering dv = 100mV
dt = 100us
In this circuit a capacitor of 220 uF is used so,
C = 220uF
Iout = 220u (0.1/100u)
Iout = 220mA
This way it can be concluded that output capacitor will respond for 220mA current change for a transient response time of 100 us.
Fig. 8: Circuit Diagram of Transient Current Compensator
Short Circuit Protection
A diode D5 is connected between the voltage input and voltage output terminals of 7809 IC so that it can prevent the external capacitor from discharging through the IC during an input short circuit. When the input is shorted then the cathode of the diode is at ground potential. The anode terminal of the diode is at high voltage since C3 is fully charged. Therefore in such a case, the diode is forward biased and all the discharging current from capacitor passes through a diode to the ground. This saves the 7809 IC from the back current. Similarly, a diode D6 is connected between the voltage input and voltage output terminals of 7909 IC which protects the IC from discharging of the capacitor C4 through the regulator when the input is shorted.
Fig. 9: Circuit Diagram of Short Circuit Protection
Testing and Precautions –
The following precautions should be taken while assembling the circuit –
• The current rating of the step-down transformer, bridge diodes and voltage regulator ICs must be greater than or equal to the required current at the output. Otherwise, it will be unable to supply the required current at the output.
• The voltage rating of the step-down transformer should be greater than the maximum required output voltage. This is due to the fact that, the 7809 and 7909 ICs take voltage drop of around 2 to 3 V. Thus input voltage must be 2V to 3V greater than the maximum output voltage and should be in the limit of the input voltage of regulator ICs.
• The capacitors used in the circuit must be of higher voltage rating than the input voltage. Otherwise, the capacitors will start leaking the current due to the excess voltage at their plates and will burst out.
• A capacitor should be used at the output of rectifier so that it can handle unwanted mains noise. Similarly Use of a capacitor at the output of the regulator is recommended for handling fast transient changes and noise at the output. The value of output capacitor depends on the deviation in the voltage, current variations and transient response time of the capacitor.
• A protection diode should always be used while using a capacitor after a voltage regulator IC, for preventing the IC from back current while discharging of the capacitor.
• For driving the high load at the output, heat sink should be mounted at the holes of the regulator. This will prevent the IC from blowing off due to heat dissipation.
• As the regulator IC can draw current up to 1A only, a fuse of 1A needs to be connected. This fuse will limit the current in the regulator up to 1A. For current above 1A, the fuse will blow off and this will cut the input supply from the circuit. This will protect the circuit and regulator ICs from current greater than 1A.
Once the circuit is assembled, it can be tested using a multimeter. Measure the output voltage at the terminals of 7809 and 7909 ICs. Then, measure the voltage outputs when loads are connected.
At 7809 regulator IC, the input voltage is 12V and the output voltage is 9.04V. With a load of 20 Ω resistance, the output voltage is read 8.03 V showing a voltage drop of 1.01 V. The output current is measured 400 mA so the power dissipation at load of 20 Ω resistance is as follow –
Pout = (Vin – Vout)*Iout
Pout = (12–8.03)*0.4
Pout = 1.58W
At 7909 regulator IC, the input voltage is -12V and output voltage is -9.18V. With a load of 20 Ω resistance, the output voltage is read -9.11 V showing a voltage drop of 0.07 V. The output current is measured 455 mA so the power dissipation at load of 20 Ω resistance is as follow –
Pout = (Vin – Vout)*Iout
Pout = (-12 – (-9.11)*0.455 (power dissipation cannot be negative)
Pout = 1.3 W
From the above tests, it can be observed that power dissipation is always less than 1.78W (internal tolerable limit of 7809 and 7909 ICs). Still, it is recommended to use a heat sink to aid cooling the IC and to increase its lifespan.
The power supply circuit designed in this project can be used to power chipsets that require a negative power supply like operational amplifiers, bi-polar amplifiers, and multi-vibrator circuits. The circuit can be used as 9V 1A power adaptor too.